What is log_{10}(1/(2*10^{-2)}) ?

The solution to log_{10}(1/(2*10^{-2)}) is -(log_{10}(2)-2)

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simplify log_{10}(1/(2*10^{-2)})

simplify

lo g_{10} (\frac{1}{2 \cdot 1 0 ^{- 2}})

- (lo g_{10} (2) - 2)

Decimal Notation

1.69897 \dots

Solution steps

lo g_{10} (\frac{1}{2 \cdot 1 0 ^{- 2}})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

lo g_{10} (\frac{1}{2 \cdot 1 0 ^{- 2}}) = lo g_{10} (1) - lo g_{10} (2 \cdot 1 0^{- 2})

= lo g_{10} (1) - lo g_{10} (1 0^{- 2} \cdot 2)

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

lo g_{10} (2 \cdot 1 0^{- 2}) = lo g_{10} (2) + lo g_{10} (1 0^{- 2})

= lo g_{10} (1) - (lo g_{10} (2) + lo g_{10} (1 0^{- 2}))

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

lo g_{10} (1 0^{- 2}) = - 2 lo g_{10} (10)

= lo g_{10} (1) - (lo g_{10} (2) - 2 lo g_{10} (10))

Apply log rule:

lo g_{a} (1) = 0

= 0 - (lo g_{10} (2) - 2 lo g_{10} (10))

2 lo g_{10} (10) = 2

= 0 - (lo g_{10} (2) - 2)

0 - (lo g_{10} (2) - 2) = - (lo g_{10} (2) - 2)

= - (lo g_{10} (2) - 2)

ln (\frac{x ^{2}}{y ^{5}})

2 lo g_{b} (z) - \frac{1}{2} lo g_{b} (y) + 4 lo g_{b} (x)

ln (- 2) - ln (- 3)

(- 4 ln ∣ 11 ∣ + 4 ln ∣ - 9 ∣) - (- 4 ln ∣ 10 ∣ + 4 ln ∣ - 10 ∣)

lo g_{3} ((x^{2} x^{5})^{3})

What is log_{10}(1/(2*10^{-2)}) ?
The solution to log_{10}(1/(2*10^{-2)}) is -(log_{10}(2)-2)